package com.wc.codeforces.二分.Easy_Demon_Problem;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.Set;
import java.util.StringTokenizer;

/**
 * @Author congge
 * @Date 2024/12/24 20:05
 * @description
 * https://codeforces.com/contest/2044/problem/F
 */
public class Main {
    /**
     * 思路：
     * suma = a[1] + a[2] + ... + a[n]
     * sumb = b[1] + b[2] + ...  +b[m]
     * all = suma * sumb
     * x == suma * sumb - a[i] * sumb - b[j] * suma + a[i] * b[j] = (suma - a[i]) * (sumb - b[j])
     * 只需要找到 x 的约数即可, 是否在两个里面都存在
     */
    static FastReader sc = new FastReader();
    static PrintWriter out = new PrintWriter(System.out);
    static int N = 200010;
    static int[] a = new int[N], b = new int[N];
    static Set<Long> A = new HashSet<>(), B = new HashSet<>();
    static int n, m, q;

    public static void main(String[] args) {
        n = sc.nextInt();
        m = sc.nextInt();
        q = sc.nextInt();
        for (int i = 1; i <= n; i++) a[i] = sc.nextInt();
        for (int i = 1; i <= m; i++) b[i] = sc.nextInt();
        long suma = 0, sumb = 0;
        for (int i = 1; i <= n; i++) suma += a[i];
        for (int i = 1; i <= m; i++) sumb += b[i];

        for (int i = 1; i <= n; i++) A.add(suma - a[i]);
        for (int i = 1; i <= m; i++) B.add(sumb - b[i]);

        while (q-- > 0) {
            int x = sc.nextInt();
            boolean res = false;
            // x为正整数的时候
            for (long i = 1; i * i <= x; i++) {
                if (x % i == 0) {
                    res |= A.contains(i) && B.contains(x / i);
                    res |= A.contains(-i) && B.contains(-x / i);
                    res |= A.contains(x / i) && B.contains(i);
                    res |= A.contains(-x / i) && B.contains(-i);
                }
            }
            // x为负数的时候
            for (long i = -1; -i * i >= x; i--) {
                if (x % i == 0) {
                    res |= A.contains(i) && B.contains(x / i);
                    res |= A.contains(-i) && B.contains(-x / i);
                    res |= A.contains(x / i) && B.contains(i);
                    res |= A.contains(-x / i) && B.contains(-i);
                }
            }
            if (res) out.println("YES");
            else out.println("NO");
        }
        out.flush();
    }
}


class FastReader {
    StringTokenizer st;
    BufferedReader br;

    FastReader() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    String nextLine() {
        String s = "";
        try {
            s = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return s;
    }

    long nextLong() {
        return Long.parseLong(next());
    }

    double nextDouble() {
        return Double.parseDouble(next());
    }

    // 是否由下一个
    boolean hasNext() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                String line = br.readLine();
                if (line == null)
                    return false;
                st = new StringTokenizer(line);
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return true;
    }
}
